The point on the unit circle with centre (0, 0); where tangents are e…

The correct answer is Option (1) → $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

unit circle with centre (0, 0)

$⇒x^2+y^2=1$  ...(1)

so $2x+2y\frac{dy}{dx}=0⇒\frac{dy}{dx}=-\frac{x}{y}$

point of equal inclination $⇒\frac{dy}{dx}=1$

$⇒-x=y$

so from (1)

$x^2+x^2=1$

$x=±\frac{1}{\sqrt{2}}$

$y=\mp\frac{1}{\sqrt{2}}$

so points → $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$